Problem
As shown in the figure, there are four circles inscribed in a right triangle, and there are large, medium, and small red circles between them.
If the diameter of the large circle is $4$ inches and the diameter of the middle circle is $2$ inches, find the diameter of the small circle.
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Solution
Let the diameters of the four circles inscribed in a right triangle be $d_1, \ d_2, \ d_3, \ and \ d_4$, respectively. Then
$$BE \ (=JK)=\sqrt{AB^2-EA^2}$$
$$=\sqrt{(\frac{d_1}{2}+\frac{d_2}{2})^2-(\frac{d_1}{2}-\frac{d_2}{2})^2}$$
$$=\sqrt{d_1 d_2}.$$
$$CF \ (=JL)=\sqrt{AC^2-CF^2}$$
$$=\sqrt{(\frac{d_1}{2}+d_2+\frac{d_3}{2})^2-(\frac{d_1}{2}-\frac{d_3}{2})^2}$$
$$=\sqrt{d_2^2+d_1 d_2+d_1 d_3+d_2 d_3}.$$
$$CG \ (=KL)=\sqrt{BC^2-GB^2}$$
$$=\sqrt{(\frac{d_2}{2}+\frac{d_3}{2})^2-(\frac{d_2}{2}-\frac{d_3}{2})^2}$$
$$=\sqrt{d_2 d_3}.$$
Here, from $JL=JK+KL$,
$$\sqrt{d_2^2+d_1 d_2+d_1 d_3+d_2 d_3}=\sqrt{d_1 d_2}+\sqrt{d_2 d_3}. \qquad [1]$$
By squaring both sides of $[1]$, we have
$$d_2^2-2d_2 \sqrt{d_1 d_3}+d_1 d_3= 0,$$
$$(d_2-\sqrt{d_1 d_3})^2=0,$$
$$∴ \ d_2=\sqrt{d_1 d_3}. \qquad[2]$$
By squaring both sides of $[2]$, we get
$$d_2^2=d_1 d_3,$$
$$\frac{d_2}{d_1} =\frac{d_3}{d_2}.$$
Thinking similarly,
$$d_3=\sqrt{d_2 d_4},$$
$$d_3^2=d_2 d_4,$$
$$\frac{d_3}{d_2}=\frac{d_4}{d_3},$$
$$∴ \ \frac{d_2}{d_1} =\frac{d_3}{d_2} =\frac{d_4}{d_3} \ (=k).$$
Therefore,
$$d_2 =kd_1,$$
$$d_3=kd_2=k^2 d_1,$$
$$d_4=kd_3=k^3 d_1.$$
Next, let the diameters of the large, middle, and small red circles be $4, \ 2,$ and $x$, respectively.
Here, since $JK=JI+KI$,
$$\sqrt{d_1 d_2}=\sqrt{d_1×4}+\sqrt{d_2×4},$$
$$\sqrt{kd_1^2}=\sqrt{4d_1}+\sqrt{4kd_1}. \qquad [3]$$
By squaring both sides of $[3]$, we have
$$kd_1=4+4k+8\sqrt{k},$$
$$∴ \ d_1=\frac{4(1+\sqrt{k})^2}{k}. \qquad [4]$$
Similarly, since $KL=RK+RL$,
$$\sqrt{d_2 d_3}=\sqrt{d_2×2}+\sqrt{d_3×2},$$
$$\sqrt{k^3 d_1^2}=\sqrt{2kd_1}+\sqrt{2k^2 d_1}. \qquad [5]$$
By squaring both sides of $[5]$, we get
$$k^2 d_1=2+2k+4\sqrt{k},$$
$$∴ \ d_1=\frac{2(1+\sqrt{k})^2}{k^2}. [6]$$
Therefore, solving the simultaneous equations $[4]$ and $[6]$, we have
$$k=\frac{1}{2} \qquad and \qquad d_1=12+8 \sqrt{2}. \qquad [7]$$
Since $RS=LR+LS$,
$$\sqrt{d_3 d_4}=\sqrt{d_3×x}+\sqrt{d_4×x},$$
$$\sqrt{k^5 d_1^2}=\sqrt{k^2 xd_1}+\sqrt{k^3 xd_1}. \qquad [8]$$
By squaring both sides of $[8]$, we get
$$k^3 d_1=x+kx+2x \sqrt{k},$$
$$∴ \ d_1=\frac{x(1+\sqrt{k})^2)}{k^3}. \qquad [9]$$
Substituting the value of $[7]$ into $[9]$, we have
$$12+8 \sqrt{2}=\frac{x (1+\sqrt{1/2})^2}{(1/2)^3},$$
$$x=\frac{12+8 \sqrt{2}}{8(1+\sqrt{1/2})^2}=1.$$
$$(Answer) \quad 1 \ inch.$$
Reference
Yoshikazu Yamakawa, ed. (1997) Okayama ken no Sangaku (Sangaku in Okayama Prefecture), p.39; pp.357-358.